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Question Number 15434 by mrW1 last updated on 10/Jun/17

Commented by mrW1 last updated on 10/Jun/17

$$\mathrm{Two}\mathrm{circles}\mathrm{with}\mathrm{radius}6\mathrm{and}10\mathrm{and}$$$$\mathrm{center}\mathrm{at}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{intersect}\mathrm{at}\mathrm{point}\mathrm{C}$$$$\mathrm{and}\mathrm{D}.\mathrm{The}\mathrm{distance}\mathrm{between}\mathrm{A}\mathrm{and}\mathrm{B}$$$$\mathrm{is}12.\mathrm{What}\mathrm{is}\mathrm{the}\max .\mathrm{length}\mathrm{of}$$$$\mathrm{segment}\mathrm{EF}\mathrm{through}\mathrm{C}?$$

Commented by ajfour last updated on 10/Jun/17

$${\left(EF\right)}_{max}=2\times \left(AB\right)$$

Commented by mrW1 last updated on 10/Jun/17

$$\mathrm{please}\mathrm{explain}\mathrm{your}\mathrm{proof}.$$

Commented by ajfour last updated on 10/Jun/17

$$PleaseviewQ.15480,Sir.$$

Commented by Tinkutara last updated on 11/Jun/17

$$\mathrm{Sir},\mathrm{will}EF\mathrm{be}\mathrm{maximum}\mathrm{when}\mathrm{it}\mathrm{is}$$$$\mathrm{parallel}\mathrm{to}AB?\mathrm{Because}\mathrm{then}\mathrm{I}\mathrm{can}$$$$\mathrm{prove}EF=2AB.$$

Commented by mrW1 last updated on 11/Jun/17

$$\mathrm{Yes},\mathrm{EF}\mathrm{is}\max .\mathrm{when}\mathrm{EF}\mathrm{is}\mathrm{parallel}$$$$\mathrm{to}\mathrm{AB}.$$$$\mathrm{We}\mathrm{can}\mathrm{prove}\mathrm{this}\mathrm{without}\mathrm{much}$$$$\mathrm{calculation}:$$$$\mathrm{We}\mathrm{look}\mathrm{at}\mathrm{the}\mathrm{triangle}\mathrm{\Delta }\mathrm{EDF}.\mathrm{We}$$$$\mathrm{know}\mathrm{no}\mathrm{matter}\mathrm{where}\mathrm{the}\mathrm{point}\mathrm{E}$$$$\mathrm{and}\mathrm{F}\mathrm{lie},\mathrm{the}\mathrm{angle}\mathrm{\angle }\mathrm{CED}\mathrm{and}\mathrm{\angle }\mathrm{CFD}$$$$\mathrm{remain}\mathrm{constant}.\mathrm{That}\mathrm{means}\mathrm{the}$$$$\mathrm{angle}\mathrm{\angle }\mathrm{EDF}\mathrm{is}\mathrm{constant}\mathrm{when}\mathrm{the}$$$$\mathrm{point}\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{change}.\mathrm{With}\mathrm{constant}$$$$\mathrm{angle}\mathrm{\angle }\mathrm{EDF}\mathrm{we}\mathrm{know}\mathrm{EF}\mathrm{is}\mathrm{maximum}$$$$\mathrm{when}\mathrm{DE}\mathrm{and}\mathrm{DF}\mathrm{are}\mathrm{maximum}.$$$$\mathrm{The}\mathrm{maximum}\mathrm{of}\mathrm{DE}\mathrm{and}\mathrm{DF}\mathrm{are}$$$$\mathrm{the}\mathrm{diameters}\mathrm{of}\mathrm{the}\mathrm{circles}.\mathrm{So}$$$$\mathrm{the}\mathrm{maximum}\mathrm{of}\mathrm{EF}\mathrm{occurs}\mathrm{when}$$$$\mathrm{DE}\mathrm{and}\mathrm{DF}\mathrm{pass}\mathrm{through}\mathrm{the}\mathrm{centers}$$$$\mathrm{of}\mathrm{the}\mathrm{circles},\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{EF}\mid \mid \mathrm{AB}\mathrm{and}$$$$\mathrm{EF}=2\mathrm{AB}.$$

Commented by mrW1 last updated on 11/Jun/17

Commented by ajfour last updated on 11/Jun/17

$$ifoundasimplerway,Sir.$$$$simplertomyearlierlengthy$$$$algebraicmethod.Kindlyviewit$$$$\left(hereinasanotheranswer\right).$$

Answered by ajfour last updated on 11/Jun/17

Commented by ajfour last updated on 11/Jun/17

$$lengthAPB=AP+PB$$$$=2(MP+PN)$$$$=2\left(MN\right)$$$$=2\left(d\cos \theta \right)$$$$=\frac{2d}{\sqrt{1+\tan {}^2\theta }}$$$$maximumlengthofAPB$$$$=2d(for\theta =0)$$$$thatiswhenitis//to{C}_1{C}_2.$$

Commented by mrW1 last updated on 11/Jun/17

$$\mathrm{you}\mathrm{have}\mathrm{shown}\mathrm{in}\mathrm{Q}15480\mathrm{that}$$$$\mathrm{the}\mathrm{maximum}\mathrm{is}\mathrm{when}\mathrm{m}=0,\mathrm{m}$$$$\mathrm{is}\mathrm{the}\mathrm{inclination}\mathrm{of}\mathrm{line}\mathrm{EF}.$$

Commented by ajfour last updated on 11/Jun/17

$$yes,butthesameexpressionfor$$$$lengthofEF=\frac{2d}{\sqrt{1+m^2}}isarrived$$$$atbythismethodwithlesseffort.$$

Commented by mrW1 last updated on 11/Jun/17

$$\mathrm{this}\mathrm{is}\mathrm{better}\mathrm{and}\mathrm{clear}!$$

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